ZOJ--1074：To the Max

最新推荐文章于 2021-01-22 21:54:59 发布

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本文探讨了在一个二维数组中寻找具有最大和的子矩形的问题，并提供了一个具体的例子及解决方案。通过定义一个数组来计算每列之和，并进一步判断这些和中哪一个对应于最大的子矩形。

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To the Max

Time Limit: 2 Seconds Memory Limit: 65536 KB

Problem

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.

As an example, the maximal sub-rectangle of the array:

0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2

is in the lower left corner:

9 2
-4 1
-1 8

and has a sum of 15.

The input consists of an N x N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N 2 integers separated by whitespace (spaces and newlines). These are the N 2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].

Output

Output the sum of the maximal sub-rectangle.

Example

Input

4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2

Output

思路：定义一个数组计算每列之和，之后判断该和是否是最大

java AC代码：

import java.util.Scanner;

public class ZOJ_1016 {
	public static void main(String[] args) {
		Scanner s=new Scanner(System.in);
		int n=s.nextInt();
		for(int i=0;i<n;i++){
			int m=s.nextInt();
			int arr[]=new int[2*m];
			for(int j=0;j<m;j++)
				arr[s.nextInt()+j]=1;
			int flag=0;
			for(int j=0;j<2*m;j++){
				if(arr[j]!=0){
					flag++;
					int num=1;
					for(int k=j;k>=0;k--){
						if(arr[k]==2)num++;
						if(arr[k]==0){
							arr[k]=2;
							break;
						}
					}
					System.out.print(num);
					if(flag<m)System.out.print(" ");
				}
			}
			System.out.println();
		}
	}
}